LouisLing
asked on
C++ Builder 4.0 - Storing bits as file
In C++ Builder 4.0, how to store bits from array of '1' and '0' to a file? I have tried calling the bits from the array and put it into a string... But when i saved it as a file, the size is large (the bits are stored as bytes, not bits).
For example, i have 4000 bits of '1' and '0' in an array. When i saved it to a file, the size is around 4000 bytes (4KBytes). What i want is that the size should be 4000 bits, but not 4000 bytes....
Please help...
Louis.
For example, i have 4000 bits of '1' and '0' in an array. When i saved it to a file, the size is around 4000 bytes (4KBytes). What i want is that the size should be 4000 bits, but not 4000 bytes....
Please help...
Louis.
ASKER CERTIFIED SOLUTION
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I would suggest that you use bitsets instead of a whole char for a 1 or a zero , this way you will also reduce your memory footprint. However only if you are willing to sacrifice some efficiency.
The STL class bitset is the best way to go about that, it also has methods for conversion to integral types also there is a to_string() operator.
Another alternative is to use C bitfields. like
struct bitfields
{
unsigned short b1 : 1;
unsigned short b2 : 1;
unsigned short b3 : 1;
....
unsigned short b8 : 1;
};
union char_bits
{
char ch;
bitfields bits;
};
this helps load and save bits faster and also has a smaller memory foot-print.
Well if you just want to do save space while saving or loading there is another alternative. Instead of writing or reading a single bit you pack/extract 32 bits into/from a single long. If your bitcount is not exactly divisible by 32 you will also need to keep track of paddings.
for e.g.
say you have 40000 bits, and you plan to save them in longs.
char *pData ; // the data that we want to write
const kSize = 40000; // total bits
int padding = kSize % sizeof(long); // padding in last write
int kCount = kSize / sizeof(long); // total writes required
// Writing
writeInt(kCount) , writeInt(padding);
for (int i=0;i < kCount; i++)
{
long temp = 0L;
for(int j=0;j<sizeof(long); j++) // pack 32 bits into a long
temp |= *(pData + i * sizeof(long) + j) ? (1 << j) : 0;
writeLongToFile(temp);
}
if(padding)
{
long temp;
for(int i=0;i<padding;i++)
temp |= *(pData + kCount * sizeof(long) + i) ? (1 << i) : 0;
}
// Reading
readInt(&kCount) , readInt(&padding);
for (int i=0;i < kCount; i++)
{
long temp = 0L;
readLongFromFile(&temp);
for(int j=0;j<sizeof(long); j++) // extract 32 bits from a long
*(pData + i * sizeof(long) + j) = (temp & (1<<j)) ? 1 : 0;
}
if(padding)
{
readlongFromfile(&temp);
for(int i=0;i<padding;i++)
*(pData + kCount * sizeof(long) + i) = (temp & (1<<j)) ? 1 : 0;
}
above is including all the bugs that it may have. but it may help you ...
The STL class bitset is the best way to go about that, it also has methods for conversion to integral types also there is a to_string() operator.
Another alternative is to use C bitfields. like
struct bitfields
{
unsigned short b1 : 1;
unsigned short b2 : 1;
unsigned short b3 : 1;
....
unsigned short b8 : 1;
};
union char_bits
{
char ch;
bitfields bits;
};
this helps load and save bits faster and also has a smaller memory foot-print.
Well if you just want to do save space while saving or loading there is another alternative. Instead of writing or reading a single bit you pack/extract 32 bits into/from a single long. If your bitcount is not exactly divisible by 32 you will also need to keep track of paddings.
for e.g.
say you have 40000 bits, and you plan to save them in longs.
char *pData ; // the data that we want to write
const kSize = 40000; // total bits
int padding = kSize % sizeof(long); // padding in last write
int kCount = kSize / sizeof(long); // total writes required
// Writing
writeInt(kCount) , writeInt(padding);
for (int i=0;i < kCount; i++)
{
long temp = 0L;
for(int j=0;j<sizeof(long); j++) // pack 32 bits into a long
temp |= *(pData + i * sizeof(long) + j) ? (1 << j) : 0;
writeLongToFile(temp);
}
if(padding)
{
long temp;
for(int i=0;i<padding;i++)
temp |= *(pData + kCount * sizeof(long) + i) ? (1 << i) : 0;
}
// Reading
readInt(&kCount) , readInt(&padding);
for (int i=0;i < kCount; i++)
{
long temp = 0L;
readLongFromFile(&temp);
for(int j=0;j<sizeof(long); j++) // extract 32 bits from a long
*(pData + i * sizeof(long) + j) = (temp & (1<<j)) ? 1 : 0;
}
if(padding)
{
readlongFromfile(&temp);
for(int i=0;i<padding;i++)
*(pData + kCount * sizeof(long) + i) = (temp & (1<<j)) ? 1 : 0;
}
above is including all the bugs that it may have. but it may help you ...
>> The STL class bitset is the best way to go about that, it also has methods
>> for conversion to integral types
Unfortunately you cannot (safely) convert data that is longer than unsigned long.
But I definitely do aggree that bitset is probably the best way to record the data (assuming the sie is constant), but you still need to get the data into a packed binary form. (Which is unfortunate, since that is how bitset probably stores it, but prevents you from seeing.)
But given the conditions of the questions, to convert an array of characters that use '1' and '0' to represent bits to an array of binary data, I would use
PackBits(char *DstPtr, const char *SrcPtr,int BitCnt)
{
while (BitCnt)
{
char Byt = 0;
for (int i = 0; i < 8 && BitCnt; ++i)
{
Byt <<= 1;
if (*SrcPtr == '1')
Byt |= 1;
--BitCnt;
}
*DstPtr++ Byt;
}
}
This code assumes that the destination array is large enough to hold the number of bytes required. That is, it must be the size of the source array (BitCnt) divided by 8. Note that the code assumes that the size of the source data is a multiple of 8. If that is not true, it can be fixed, at soem additional cost in complexity, but it woudl be rare for there to be cases where that would not be true.
You woudl use it like
char Result[2];
char Data[17] = '1011001010110101';
PackBits(Result,Data,16);
>> for conversion to integral types
Unfortunately you cannot (safely) convert data that is longer than unsigned long.
But I definitely do aggree that bitset is probably the best way to record the data (assuming the sie is constant), but you still need to get the data into a packed binary form. (Which is unfortunate, since that is how bitset probably stores it, but prevents you from seeing.)
But given the conditions of the questions, to convert an array of characters that use '1' and '0' to represent bits to an array of binary data, I would use
PackBits(char *DstPtr, const char *SrcPtr,int BitCnt)
{
while (BitCnt)
{
char Byt = 0;
for (int i = 0; i < 8 && BitCnt; ++i)
{
Byt <<= 1;
if (*SrcPtr == '1')
Byt |= 1;
--BitCnt;
}
*DstPtr++ Byt;
}
}
This code assumes that the destination array is large enough to hold the number of bytes required. That is, it must be the size of the source array (BitCnt) divided by 8. Note that the code assumes that the size of the source data is a multiple of 8. If that is not true, it can be fixed, at soem additional cost in complexity, but it woudl be rare for there to be cases where that would not be true.
You woudl use it like
char Result[2];
char Data[17] = '1011001010110101';
PackBits(Result,Data,16);
Opps
*DstPtr++ Byt;
should be
*DstPtr++ = Byt;
*DstPtr++ Byt;
should be
*DstPtr++ = Byt;
ASKER
I have got the answer... The working steps is like this:
FILE *fp;
fp = fopen("TEST.txt", "wb");
int counting = 0;
while(counting < KeyIdx){
Byte mask = 0x00;
for(int j=0; j<8; j++){
mask |= (SKey[counting]<<j);
++counting;
if(counting == KeyIdx){
break;
}
}
fwrite (&mask, 1 , 1 , fp);
}
fclose(fp);
Thank you for the help...
FILE *fp;
fp = fopen("TEST.txt", "wb");
int counting = 0;
while(counting < KeyIdx){
Byte mask = 0x00;
for(int j=0; j<8; j++){
mask |= (SKey[counting]<<j);
++counting;
if(counting == KeyIdx){
break;
}
}
fwrite (&mask, 1 , 1 , fp);
}
fclose(fp);
Thank you for the help...
Hmmm .... i hope you did read all the comments
this is a working c++ program that should answer your needs , i hope that you will appreciate the time that i spend on this , and that you will find it helpfull...
#include <iostream>
#include <process.h>
using namespace std;
void SaveBitStream(char *sFileName,BYTE *bitArray)
{
FILE *bitstream = fopen(sFileName,"w+b");
// number of bytes that will cover the bits
// in buffer
int nNumBytes =
((sizeof(bitArray)/sizeof(
// loop on the byte array
for (int i=0;bitArray[i]!=-1;++i)
{
BYTE CurrByte=0;
// calculate the bit value for this byte for (int j=i,k=0;((k<8)&&
(bitArray[j]! =0xFF));++j,++k)
{
CurrByte+=
(j==7?(bitArray[j]*1):(bit
}
// save the byte to the file
fputc(CurrByte,bitstream);
// time to stop?
if (bitArray[j]==0xFF)
{
break;
}
i=j-1;
}
fclose(bitstream);
}
void ReadBitStream(char *sFileName,BYTE *bitArray)
{
FILE *bitstream = fopen(sFileName,"rb");
// get the number of bytes to read
fseek(bitstream,SEEK_END,0
int nNumBytes = ftell(bitstream);
fseek(bitstream,SEEK_SET,0
// read the bits from file , they are bytes
// so obtain thier setting: 0 or 1
int inx = 0;
for (int i=0;i<=nNumBytes;++i)
{
BYTE Byte = fgetc(bitstream);
for (int j=0;j<8;++j,++inx)
{
bitArray[inx] =
(j==7?(Byte&1):((Byte&(2<<
}
}
// put an end mark in the bit array
bitArray[inx] = 0xFF;
fclose(bitstream);
}
void main()
{
// lets assume that we are saving this bit
// array , if the number of bits is not
// a multiplication of 8 , then we put 0 till we
// obtain that!
// the right byte in each segment of 8 bytes that
//represent one byte (8 bits) is the
// most significant bit and the left byte is the
// least significant bit!
BYTE SaveBuf[] = {
/*MSB*/ 1,1,1,1,1,1,1,1, /*LSB*/ // BYTE 1
0,1,0,1,0,0,0,1, // BYTE 2
0,1,-1}; // BYTE 3
// before saving
cout<<"We are saving this :\n\n";
for (int i=0;SaveBuf[i]!=0xFF;++i)
{
cout<<((char)(SaveBuf[i]+'
}
cout<<endl;
// save to file
SaveBitStream("bitstream",
// read from file
BYTE ReadBuf[512] = {0};
ReadBitStream("bitstream",
cout<<"\n\nAfter Read :\n\n";
for (i=0;ReadBuf[i]!=0xFF;++i)
{
cout<<((char)(ReadBuf[i]+'
}
cout<<endl;
system("pause");
}
OL this question is a year old!
These questions you are locking were answered months ago. The questioner has forgotten about them and has not bothered to award points to the experts that helped him/her. But now you awre comming in an locking them. You don't deserve the points, the work was done months or years ago.
These questions you are locking were answered months ago. The questioner has forgotten about them and has not bothered to award points to the experts that helped him/her. But now you awre comming in an locking them. You don't deserve the points, the work was done months or years ago.
Answer rejected, please guide me on the fair outcome here. Asker clearly chose to abandon this question versus award points for your excellent assistance. Very sad, indeed.
EE Moderator
EE Moderator
Its hard to choose between zoppo and ambience's answer, but zoppo's approach works (but is less elegant) and was first.
Thank you very much for the advice, nietod, I very much appreciate it. The first correct answer was accepted. The other option may have been to change this value to 200 and offer ambience 100 for the additional enhancement. If anyone feels I've decided incorrectly, please let me know and I'll make it right. 300 max per question, and that was the question value.
Thanks again,
EE Moderator
Thanks again,
EE Moderator
Well, thanks a lot, Moondancer and nietod ...
I would let ambience decide about this ... I wouldn't have any problems
with sharing the points.
Have a nice day,
regards,
ZOPPO
I would let ambience decide about this ... I wouldn't have any problems
with sharing the points.
Have a nice day,
regards,
ZOPPO
ASKER
for ( DWORD j = 0; j < s; j++ )
{
if ( ( m_pBuf[i] & ( 1 << j ) ) != 0 )
pData[i * BITS_PER_DWORD + j] = 1;
else
pData[i * BITS_PER_DWORD + j] = 0;
}
}
ermm... where is the 'i' in this code?